Example of Condorcet Non-additivity

Consider the following example.

District I
#voters Their Vote
3 A>B>C>D
2 B>C>A>D
2 C>A>B>D
District II
#voters Their Vote
3 A>D>C>B
2 D>C>A>B
2 C>A>D>B

In district I, A wins (in most Condorcet methods, anyhow). In district II, (same as district I but the roles of B and D are swapped), A also wins. But in the combined 2-district country, C is the Condorcet winner (beats A by 8:6, beats B and D by 9:5).

Consequently, some Condorcet methods cannot be "counted in Precincts." For example, the Smith,IRV method invented by Woodall is Condorcet but I see no efficient way to count it in precincts if there are a large number of candidates (say 100 candidates, if you want something concrete to think about – see any feasible way to count it other than a centralized count? I don't).

However, the Condorcet methods that depend only on the matrix of "pairwise totals" can be counted in precincts (despite the nonadditivity paradox illustrated above!) in the sense that each precinct can find and publish its pairwise matrix and the summed matrix will be used for the whole country.

Here is a second example now with only 3 candidates:

District I
#voters Their Vote
30 A>B>C
20 B>C>A
20 C>A>B
20 B>A>C
District II
#voters Their Vote
21 A>C>B
20 C>B>A
20 B>A>C

A wins in district I with every Condorcet method since A beats B pairwise 50:40 and also beats C pairwise 50:40.

In district II, the situation is completely symmetric between A, B, and C except for a single extra ballot of type "A>C>B." Therefore, A wins in every Condorcet method ever seriously proposed, although A is not a "beats-all winner" because there is a cycle.

But in the combined country, B wins with every Condorcet method since B beats A pairwise 80:71 and also beats C pairwise 90:61.

For those who cannot get enough, here is a third example again with only 3 candidates:

District I
#voters Their Vote
3 A>B>C
4 B>C>A
3 C>A>B
District II
#voters Their Vote
2 A>B>C
3 B>A>C

In district I, the situation is completely symmetric between A, B, and C except for a single extra ballot of type "B>C>A." Therefore, B wins in every Condorcet method ever seriously proposed, although B is not a "beats-all winner" because there is a cycle.

In district II, B is ranked top by 60% of the voters hence wins in in every Condorcet method. (Also wins with Borda and virtually every other voting method.)

But in the combined country, A wins with every Condorcet method since A beats B pairwise 8:7 and also beats C pairwise 8:7.


Analogous non-additivity problem for IRV system

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